Mathematical Methods for Physicists: A concise introduction - Site Map

FUNCTIONS OF A COMPLEX VARIABLE
Example 6.14
Evaluate the integral H C ez dz=…z 2 ‡ 1†, ifC is a circle of unit radius with center at
(a) z ˆ i and (b) z ˆi.
Solution:
(a) We ®rst rewrite the integral in the form
I
e z dz
z ‡ i z i ;
C
then we see that f …z† ˆe z =…z ‡ i† and z 0 ˆ i. Moreover, the function f …z† is
analytic everywhere within and on the given circle of unit radius around z ˆ i.
By Cauchy's integral formula we have
I
C
e z dz
z ‡ i z i
ei
ˆ 2if …i† ˆ2i ˆ …cos 1 ‡ i sin 1†:
2i
(b) We ®nd z 0 ˆi and f …z† ˆe z =…z i†. Cauchy's integral formula gives
I
e z dz ˆ…cos 1 i sin 1†:
z i z ‡ i
C
Cauchy's integral formula for higher derivatives
Using Cauchy's integral formula, we can show that an analytic function f …z† has
derivatives of all orders given by the following formula:
f …n† …z 0 †ˆ n! I
f …z†dz
2i C …z z 0 † n‡1;
…6:28†
where C is any simple closed curve around z 0 and f …z† is analytic on and inside C.
Note that this formula implies that each derivative of f …z† is itself analytic, since it
possesses a derivative.
We now prove the formula (6.28) by induction on n. That is, we ®rst prove the
formula for n ˆ 1:
f 0 …z 0 †ˆ 1 I
f …z†dz
2i C …z z 0 † 2:
As shown in Fig. 6.12, both z 0 and z 0 ‡ h lie in R, and
f 0 f …z
…z 0 †ˆlim 0 ‡ h†f …z 0 †
:
h!0 h
Using Cauchy's integral formula we obtain
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